S - Block MCQ Quiz - Objective Question with Answer for S - Block - Download Free PDF

Explanation:

Carbon dioxide\( (\text{CO}_2)\) gas is bubbled through a slaked lime \((\text{Ca(OH)}_2)\) solution.

Reaction Explanation

• When carbon dioxide gas is bubbled through a solution of slaked lime (aqueous calcium hydroxide), the following reaction occurs:
  • \(\text{Ca(OH)}_2(aq) + \text{CO}_2(g) \rightarrow \text{CaCO}_3(s) + \text{H}_2\text{O}(l)\)
• This reaction results in the formation of calcium carbonate \((\text{CaCO}_3)\) as a solid precipitate and water \((\text{H}_2\text{O}).\)

Conclusion:

The compound formed when carbon dioxide gas is bubbled through slaked lime solution is  \((\text{CaCO}_3)\)

The correct answer is silver.

Key Points

• Silver is a chemical element with the symbol Ag and atomic number 47.
• It is a soft, white, lustrous transition metal, known for its high electrical and thermal conductivity.
• Silver is widely used in industrial applications, jewelry, and as currency in the form of coins and bullion.
• It is also utilized in photography, mirrors, and as an antimicrobial agent in various products.

Additional Information

• Fluorine
  • Fluorine is a chemical element with the symbol F and atomic number 9.
  • It is a highly reactive, pale yellow gas at room temperature and is the most electronegative element.
  • Fluorine is used in the production of fluorocarbons, toothpaste, and Teflon.
• Phosphorus
  • Phosphorus is a chemical element with the symbol P and atomic number 15.
  • It exists in several forms, including white, red, and black phosphorus.
  • Phosphorus is essential for life and is a key component of DNA, RNA, and ATP.
• Sulphur
  • Sulphur is a chemical element with the symbol S and atomic number 16.
  • It is a yellow, non-metallic element that is essential for all living organisms.
  • Sulphur is used in the production of fertilizers, chemicals, and in vulcanization of rubber.

CONCEPT:

Thermal Stability of Hydroxides

• Thermal stability of hydroxides refers to the ability of a hydroxide to remain stable (not decompose) when heated.
• For group 2 hydroxides (alkaline earth metal hydroxides), the thermal stability increases down the group.
• This is due to the increasing ionic size and decreasing lattice energy as we move down the group, which makes the hydroxides more thermally stable.

EXPLANATION:

• In the given options, we are comparing the thermal stability of Mg(OH)₂, Ca(OH)₂, Sr(OH)₂, and Ba(OH)₂.
• As we go down the group in the periodic table:
  • Mg(OH)₂ (Magnesium hydroxide) is the least thermally stable.
  • Ca(OH)₂ (Calcium hydroxide) is more stable than Mg(OH)₂.
  • Sr(OH)₂ (Strontium hydroxide) is more stable than Ca(OH)₂.
  • Ba(OH)₂ (Barium hydroxide) is the most thermally stable.
• Therefore, the correct order of thermal stability is:
  • Mg(OH)₂ < Ca(OH)₂ < Sr(OH)₂ < Ba(OH)₂

Therefore, the correct order of thermal stability of hydroxides is Mg(OH)₂ < Ca(OH)₂ < Sr(OH)₂ < Ba(OH)₂, which corresponds to option 3.

CONCEPT:

Reaction of hot and concentrated NaOH with dichlorine (Cl₂)

• When dichlorine reacts with hot and concentrated sodium hydroxide (NaOH), a disproportionation reaction occurs. In this reaction, chlorine undergoes both oxidation and reduction to form different products.
• The balanced chemical equation for this reaction is:

  3 Cl₂ + 6 NaOH → 5 NaCl + NaClO₃ + 3 H₂O

EXPLANATION:

• In the given reaction:

  3 Cl₂ + 6 NaOH → 5 NaCl + NaClO₃ + 3 H₂O

  • Chlorine (Cl₂) is reduced to form sodium chloride (NaCl).
  • Chlorine (Cl₂) is also oxidized to form sodium chlorate (NaClO₃).
• Thus, the products obtained are:
  • Sodium chloride (NaCl)
  • Sodium chlorate (NaClO₃)
  • Water (H₂O)

Therefore, the correct answer is option 3: NaCl + NaClO₃ + H₂O.

Explanation:

Given:

The question asks which of the following compounds is thermodynamically the most stable.

The thermal stability of carbonates increases down the group. Therefore, the order of thermal stability of carbonates of alkaline earth metals is:

BaCO₃ > CaCO₃ > MgCO₃ > BeCO₃

Thus, BaCO₃ is thermodynamically the most stable and BeCO₃ is the least stable.

∴ The correct answer is BaCO₃.

The correct answer is Alkali metal group.

Key Points

• Alkali Metals:
  • Group 1 of the periodic table consists of namely, lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs), and francium (Fr). 
• Properties of Alkali metals:
  • They have high thermal and electrical conductivity, lustre, ductility, and malleability.
  • All these elements react with water to form hydrogen gas.
  • Each of these, alkali metal atoms has a single electron in its outermost shell. Hence, their valency is 1.

The correct answer is Only I.

Key Points

• Lithium upon reaction with excess air usually forms lithium oxide but lithium peroxide is also possible to form.
• But we know that peroxides upon hydrolysis give hydrogen peroxide and metal hydroxide.
• The reaction between lithium and oxygen is 4Li + O₂→2Li₂O.
• The alkali metal halides exist as colourless crystalline solids, although as finely ground powders appear white.
• They melt at high temperature, usually several hundred degrees to colorless liquids.
• Their high melting point reflects their high lattice energy.

Additional Information

• Alkali Metal:
  • Alkali metal form a part of the leftmost group, i.e. group 1 of the ultramodern periodic table.
  • It contains six rudiments, which are collaboratively called the Lithium family.
  • These essence are largely reactive essence of the periodic table due to the presence of one electron in their remotest shell, called the valence electron.
• Alkaline Earth Metal:
  • Elements whose atoms have their s-subshell filled with their two valence electrons are called alkaline earth metals.
  • Their general electronic configuration is [Noble gas] ns2.
  • They occupy the second column of the periodic table and so-called as group two metals also.

• Calcium carbonate (CaCO₃) is a chemical compound, formed by three main elements: carbon, oxygen, and calcium. 
• Calcium carbonate is used to prevent calcium deficiency.
• Gunpowder or Black Powder is a mix of three different components.

Concept:

Lithium nitrate decomposes into its oxide on heating. Upon thermal decomposition, LiNO₃ gives lithium oxide (Li₂ O), nitrogen dioxide (NO₂), and oxygen (O₂). LiNO₃ on heating gives reddish-brown fumes of NO₂.

\(4{\rm{LiN}}{{\rm{O}}_3}\mathop \to \limits^{\rm{\Delta }} 2{\rm{L}}{{\rm{i}}_2}{\rm{O}} + 4{\rm{N}}{{\rm{O}}_2} + {{\rm{O}}_2}\)

Other groups of nitrates decompose differently, forming the nitrate salt and oxygen. Because of its relatively small size, the lithium cation is very polarizing, which favors the formation of the oxide.

The other statements are true about lithium.

Concept:

The temporary hardness is caused by bicarbonates of calcium and magnesium. On boiling following changes occurs,

On boiling, the bicarbonates salt of magnesium Mg(HCO₃)₂, the solution produces the magnesium hydroxide to decompose and yielding carbon dioxide and hydrogen.

\({\rm{Mg}}{\left( {{\rm{HC}}{{\rm{O}}_3}} \right)_2}\left( {{\rm{aq}}} \right)\mathop \to \limits^{{\rm{\;boiling\;}}} {\rm{Mg}}{({\rm{OH}})_2} + {\rm{C}}{{\rm{O}}_2} \uparrow + {{\rm{H}}_2} \downarrow \)

Calcium hydrogen carbonate on boiling will produce calcium carbonate with carbon dioxide and hydrogen.

\({\rm{Ca}}{\left( {{\rm{HC}}{{\rm{O}}_3}} \right)_2}\left( {{\rm{aq}}} \right)\mathop \to \limits^{{\rm{\;boiling\;}}} {\rm{CaC}}{{\rm{O}}_3}{\rm{\;}} + {\rm{C}}{{\rm{O}}_2} \uparrow + {{\rm{H}}_2} \downarrow \)

X = Mg(HCO₃)₂

Concept:

• Non metals exist in solid liquid and gases state.
• They are all placed in p-block towards right hand side of periodic table from group no 13 to 18.

Explanation:

• Talking about halogen they are placed in group 17
• Member of this family include Fluorine(F) Chlorine (Cl), Bromine(Br), Iodine(I), Astatine(At), and Tennessine(Ts).
• Among these only Br exist in liquid state and F, Cl, exist in gases state while Iodine is a solid.
• As the atomic no. increases the Vander waal force of attraction increases. 
• In case of Br intermolecular forces are strong enough so that it does not evaporate.
• Bromine forms diatomic molecules and Van der Waals interactions are sufficiently strong.
• Also Br is more reactive than Iodine but less reactive than Cl.
• After Br, as we move down the group iodine occur as solid because Vander Waal forces become more stronger than Br and hence physical state become dense.
• We conclude that down the group the physical state become more denser therefore we observed that F, Cl are gases, Br is liquid and I is solid.
• Among other non-metals like Carbon, Sulphur, Phosphorous, etc are solid whereas Oxygen, Nitrogen, are gases.

​

Concept:

The reaction is given below:

Na₂ CO₃.10H₂O ⟶ Na₂ CO₃.H₂O + 9H₂O

            (X)                    (Y)

Na₂ CO₃.H₂O ⟶ Na₂ CO₃ + H₂O

 (Y)                   (Z)

Here, X represents Washing soda

Y represents Washing soda (monohydrated)

and Z represents Soda ash

When washing soda is heated which undergoes dehydrogenation to form washing soda. Then again heated above 373 K to form sodium carbonate powder with very less water content that is soda ash.

CONCEPT:

Concentration of Ores Using Cyanide

• In metallurgy, various methods are used to concentrate ores, which involve separating the valuable minerals from the gangue (the non-valuable minerals).
• Cyanide is commonly used in the froth flotation process to selectively depress certain sulfide minerals.
• Cyanide particularly affects the concentration of galena (PbS) and sphalerite (ZnS) ores.

EXPLANATION:

• Sphalerite (ZnS): Cyanide is used as a depressant to prevent sphalerite from forming froth in the froth flotation process. This is done to selectively concentrate galena when both ores are present. Sphalerite itself is not concentrated by cyanide, but its separation is facilitated in the presence of galena.
• Malachite (Cu₂(OH)₂CO₃): This is an ore of copper and is typically not concentrated using cyanide.
• Calamine (ZnCO₃): This is a zinc ore and is also not concentrated using cyanide.
• Siderite (FeCO₃): This is an iron ore and is not concentrated using cyanide.

Given that the cyanide is used to depress sphalerite in the presence of galena, the correct interpretation in the context of the provided explanation is that sphalerite is involved in a process where cyanide is used.

The correct answer is: 1) Sphalerite

Concept:

Nitrogen in the form of N₂ gas makes up around 80% of air, and Oxygen in the form of O₂ makes up around 20%. There is a fraction of a percent of CO₂ as well as traces of other gasses. The largest component of air, by volume, is nitrogen. Thus, we need to consider both N₂ and O₂.

Hence, magnesium reacts with both nitrogen and oxygen to give the following compounds:

\(Mg + {O_2}\mathop {\xrightarrow[]{\Delta}}\limits^{\;\;\;\;\;\;\;\;\;\;\;\;\;} MgO\)

\(Mg + {N_2}\mathop {\xrightarrow[]{\Delta}}\limits^{\;\;\;\;\;\;\;\;\;\;\;\;\;} Mg_3N_2\)

Thus, Mg burns in air and produces a mixture of nitride and oxide.

The correct answer is option 4.

Concept:

• Group 1 elements in the periodic table are alkali metals. They are francium (Fr), cesium (Cs), lithium (Li), sodium (Na), potassium (K), and rubidium (Rb).
• These metals are extremely reactive because they only have one electron in their outermost energy level.
• The capacity of alkali metals to react with oxygen to generate various oxides, including superoxides, peroxides, and regular oxides, is one of their special qualities.

Explanation:

1. Normal Oxides (O₂^-): The oxidation state of oxygen in these oxides is -2. For example, 4Li(s) + O₂(g) -> 2Li₂O(s).
2. Peroxides (O₂^2-): The oxidation state of each oxygen atom in a peroxide is -1. For example, 2Na(s) + O₂(g) -> Na₂O₂(s).
3. Superoxides (O₂^-):  In superoxides, the oxygen has an oxidation state of -1/2. For example, K(s) + O₂(g) -> KO₂(s).

Additional Information

• The various alkali metals form different oxides due to their differing reactivities and the size of the metal ions.
• As the atomic size increases down the group, the stability of the superoxide ion increases leading to the formation of superoxides for the larger alkali metals like potassium, rubidium and cesium.
• The small size of lithium and sodium make the formation of superoxides energetically unfavorable - they preferentially form the more compact oxides or peroxides.

Hence , the correct answer is option 4.